To find the local minima of the function f(x) = sin(x) + 1/2 cos(x) where 0 < x < pi/2, we need to find the critical points of the function in the given interval.
To find the critical points, we need to find where the derivative of the function is equal to zero or does not exist.
First, let's find the derivative of f(x).
f'(x) = cos(x) - 1/2 sin(x)
Next, we set this derivative equal to zero and solve for x:
cos(x) - 1/2 sin(x) = 0
2cos(x) - sin(x) = 0
To solve this equation, we can square both sides:
(2cos(x) - sin(x))^2 = 0
Expanding, we get:
4cos^2(x) - 4cos(x)sin(x) + sin^2(x) = 0
Using the trigonometric identity cos^2(x) + sin^2(x) = 1, we can simplify the equation:
4(1 - sin^2(x)) - 4cos(x)sin(x) + sin^2(x) = 0
4 - 4sin^2(x) - 4cos(x)sin(x) + sin^2(x) = 0
5sin^2(x) + 4cos(x)sin(x) - 4 = 0
Now, let's solve this quadratic equation for sin(x):
5sin^2(x) + 4cos(x)sin(x) - 4 = 0
Since the range of sin(x) is [-1, 1], we know that sin(x) cannot be equal to 4/5 or -4/5. Therefore, the quadratic equation does not have any real solutions for sin(x).
Since the derivative does not equal zero in the given interval, there are no critical points and therefore no local minima for the function f(x) = sin(x) + 1/2 cos(x) in the interval 0 < x < pi/2.
The point of local minima of f(x) =sin x +1/2 cos x
where 0 < x <pi/2
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